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Year 2023
September
Week # 2
Expire on 07 Oct 2023
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FAQs
September
Week # 3 Quiz 3
Q1.
Find the output voltage in the following circuit:
\(250\:mV\)
\(-25\:mV\)
\(25\:mV\)
\(-250\:mV\)
Q2.
In voltage divider bias, \(V_{CC} = 25\: V;\: R_1 = 10\: k\Omega;\: R_2 = 3.3 \: k\Omega; \:R_C = 2 \: k\Omega;\) and \(R_E =1\: k\Omega\). What is the collector current?
6.2 mA
5.5 mA
4.5 mA
3.6 mA
Q3.
Let a typical audio amplifier have a transfer function \(G(j\omega) = 10 / (1 + j\omega/100)\)
Male voice will be more attenuated than that of the female
Female voice will be more attenuated than that of the male
Both male and female voice will be distorted by equal amount
Only the female voice will be distorted
Q4.
In voltage divider bias, \(V_{CC} = 20\: V;\: R_1 = 10\: k\Omega;\: R_2 = 3.3 \: k\Omega; \:R_C = 3 \: k\Omega;\) and \(R_E =1.5\: k\Omega\). What is the emitter voltage?
4.96 V
5.66 V
3.8 V
4.26 V
Q5.
Choose the correct match for input resistance of various amplifier configurations shown below: $$\begin{array} {|r|r|}\hline CB : Common\: Base & LO : Low \\ \hline CC : Common \:Collector & MO : Moderate \\ \hline CE : Common\: Emitter & HI : High \\ \hline \end{array}$$
\(CB-LO, \:CC-MO, \:CE-HI\)
\(CB-LO, \:CC-HI,\: CE-MO\)
\(CB-MO,\: CC-HI,\: CE-LO\)
\(CB-HI,\: CC-LO,\: CE-MO\)
Q6.
In a transistor, signal is transferred from
high resistance to low resistance
low resistance to high resistance
high resistance to high resistance
low resistance to low resistance
Q7.
For a transistor connected in common base connection, collector current is 0.95 mA and base current is 0.05 mA. Find the value of \(\alpha\):
1.00
0.5
0.95
0.05
Q8.
A non inverting amplifier is nulled at 25 ºC with a gain of 100. Calculate the output voltage when the temperature rises to 50 ºC for an offset drift voltage of 0.15 mV/ ºC.
3.75mV
2.75mV
37.5mV
375mV
Q9.
An amplifier without feedback has a voltage gain of 60, input resistance of \(5\:K\Omega\) & Output resistance of \(2\:K\Omega\). The input resistance of the current-shunt negative feedback amplifier using the above amplifier with a feedback factor of 0.2 is
\(\frac{5}{12}\: K\Omega\)
\(\frac{2}{12}\: K\Omega\)
\(12\: K\Omega\)
\(2\: K\Omega\)
Q10.
The positive clipper can be easily converted into a negative clipper by
Reversing diode and changing the polarity of reference voltage
Reversing diode only
Changing the polarity
None of these
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